Giải bài tập 3.5 trang 32 SBT toán 10 tập 1 kết nối
Bài tập 3.5. Chứng minh rằng:
a) $sin^{4}\alpha + cos^{4}\alpha = 1 - 2sin^{2}\alpha cos^{2}\alpha$;
b) $sin^{6}\alpha + cos^{6}\alpha = 1 - 3sin^{2}\alpha cos^{2}\alpha$;
c*) $\sqrt{sin^{4}\alpha + 6cos^{2}\alpha + 3} + \sqrt{cos^{4}\alpha + 4sin^{2}\alpha}$ = 4.
Trả lời:
a) $sin^{4}\alpha + cos^{4}\alpha = 1 - 2sin^{2}\alpha cos^{2}\alpha$
Xét $sin^{4}\alpha + cos^{4}\alpha$
= $(sin^{2}\alpha)^{2} + (cos^{2}\alpha)^{2}$
= $sin^{2}\alpha + 2sin^{2}\alpha cos^{2}\alpha + cos^{2}\alpha - 2sin^{2}\alpha cos^{2}\alpha$
= $(sin^{2}\alpha + 2sin^{2}\alpha cos^{2}\alpha + cos^{2}\alpha) - 2sin^{2}\alpha cos^{2}\alpha$
= $(sin^{2}\alpha + cos^{2}\alpha)^{2} - 2sin^{2}\alpha cos^{2}\alpha$
= 1 - $2sin^{2}\alpha cos^{2}\alpha$
b) $sin^{6}\alpha + cos^{6}\alpha = 1 - 3sin^{2}\alpha cos^{2}\alpha$
Xét $sin^{6}\alpha + cos^{6}\alpha$
= $(sin^{2}\alpha)^{3} + (cos^{2}\alpha)^{3}$
= $(sin^{2}\alpha + cos^{2}\alpha).(sin^{4}\alpha - sin^{2}\alpha cos^{2}+cos^{4}\alpha)$
= $(sin^{2}\alpha + cos^{2}\alpha).(sin^{4}\alpha + 2sin^{2}\alpha cos^{2}\alpha + cos^{4}\alpha - 3sin^{2}\alpha cos^{2}\alpha)$
= 1.$(sin^{4}\alpha + 2sin^{2}\alpha cos^{2}\alpha + cos^{4}\alpha - 3sin^{2}\alpha cos^{2}\alpha)$
= $(sin^{2}\alpha + cos^{2}\alpha)^{2} - 3sin^{2}\alpha cos^{2}\alpha$
= 1 - $3sin^{2}\alpha cos^{2}\alpha$
c*) $\sqrt{sin^{4}\alpha + 6cos^{2}\alpha + 3} + \sqrt{cos^{4}\alpha + 4sin^{2}\alpha}$ = 4
Xét $\sqrt{sin^{4}\alpha + 6cos^{2}\alpha + 3} + \sqrt{cos^{4}\alpha + 4sin^{2}\alpha}$
= $\sqrt{(1-cos^{2}\alpha)^{2} + 6cos^{2}\alpha + 3} + \sqrt{(cos^{2})^{2} + 4sin^{2}\alpha}$
= $\sqrt{4 + 4cos^{2}\alpha + cos^{4}\alpha} + \sqrt{1 + 2sin^{2}\alpha + sin^{4}\alpha}$
= $(2 + cos^{2}\alpha) + (1 + sin^{2}\alpha)$
= 4
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